My Physics Teacher didn't tell me how to do this

AI Thread Summary
The discussion centers on a physics assignment involving pendulums and harmonic motion, where key concepts include period, frequency, and the relationship between length and period. The formula for the period of a pendulum is provided as T = 2π√(l/g), with frequency defined as the inverse of the period. The user seeks assistance in solving specific problems related to pendulum length and period, expressing confusion about potential connections to centripetal force. Responses clarify the formulas needed and suggest rechecking calculations for accuracy. The conversation highlights the challenges of understanding the material and the need for clear guidance from the teacher.
Earth_kissed
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he gave me an assignment about Pendulums and Harmonic motion. He gave us notes on vocab words we would need to know like; Bob, Equilibrium, period, frequency, and oscilation. he also told us that Period is unaffected by height of drop. That Length is directly proportional to Period. and Length is inversly proportional to frequency. and mass doesn't affect anything... However these are the problems he gave us:

1. What is the length of a pendulum with a period of 2 seconds?

2. What is the period of a pendulum with a length of 67 meters?

3. What is the same Pendulum's frequency?

4. How long would the pendulum be if you doubled the period?

5. How long would the pendulum be if you doubled the frequency?

All I need is a formula or equation... but if you want to give me the procedure as well that would be helpfull.


Thank you,

Laura
 
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OK,

You must not have a very good textbook. This equation for the period of a simple pendulum should be useful to you:

T = 2\pi\sqrt{\frac{l}{g}}
 
T = 2 \pi \sqrt{\frac{l}{g}}

T is the time or period in seconds.
l is the length in meters.
g is the acceleration of gravity in meters/second^2

Frequency is the inverse of time so f = \frac{1}{T}

To solve for l, you would need to arrange the formula accordingly.

Square both sides to get rid of the radical:

T^2 = 4 \pi ^2 \frac{l}{g}

Arrange to solve for l:

l = \frac{T^2 g}{4 \pi ^2}
 
thank you very much :)

these the answers I got... hope they arn't wrong!

1. .99m

2. 1.64s

3. .61 oscilatons per second

4. 2.68m

5. .17m

now I'm a little worried because my teacher did go over Centripetal force... maybe with these questions he was talking about a circular motion... because he absolutly did not give us that equation. But he did go over all the equations for Cenripetal force... I'm confused...
 
The answer for question 1 is correct. Recheck your calculation for question 2.

Are questions 3 - 5 referring to the pendulum in question 1 or question 2?
 
Originally posted by Earth_kissed
thank you very much :)

now I'm a little worried because my teacher did go over Centripetal force... maybe with these questions he was talking about a circular motion... because he absolutly did not give us that equation. But he did go over all the equations for Cenripetal force... I'm confused...

Oh. Well the formula we gave is derived from other equations dealing with angular motions and forces... Exactly what info were you given?
 
I have to log off now but I'm sure someone else will be around to help out. Good luck!
 
wow, my teacher did the EXACT same thing to me, but I had to find the conditions/problems in the experiment myself:frown: . Oh well.
 
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